## The mathematical repository, Volume 1 |

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ANALYSIS answered arch ARTICLE base bisected cafe centre circle given circumference demit Demon demonstration described diameter distance draw drawn perpendiculars drawn right lines ellipsis equal to twice equation equilateral figure fame ratio fides fluxion Frustums given by position given points given space hence inscribed intersecting join KCML latus rectum let an equilateral lines be drawn lines drawn Lowry meet the circle number of given parabola parallel pendicular perpendicular perpendiculars drawn Plate points found Prop proposition QUESTION radius rect rectangle contained right angle right lines given segment semicircle semidiameter sides spherical triangle square of XY Swale tangent THEO THEOREM theref Thornoby twice the multiple twice the sum vertical angle vibration wherefore

### Popular passages

Page 19 - If the circumference of a circle be divided into any number of equal parts, the chords joining the successive points of division form a regular polygon inscribed in the circle ; and the tangents drawn at the points of division form a regular polygon circumscribed about the circle.

Page 148 - I say that EC : CD : : EH : HG. PRoP. IX. If in AB the diameter of a circle be taken any point C, and CD be drawn meeting the circumference in D and E, and from the point D be drawn DF perpendicular to CD, which meets the diameter AB in Fand the circumference in G, then I say that DC : CE : : DF : FG.

Page 287 - Figure circumscrib'd about a Circle " and from any point in the Circumference of the Circle let there be " drawn perpendiculars to the sides of the Figure, twice the...

Page 286 - LDK is equal to the square of DE. PROP. XXII. If in AB the diameter of a circle be taken two points C and D such that AC : CB : : AD : DB, and D be without the circle, and DE be perpendicular to AB, and through C be drawn any line meeting the circle in G and H, and the line DE in K, and GL touch the circle in G, and meet DE in L; then I say the rectangle LDK is equal to the rectangle ADB. 3H 2 *nE = *GE = GEE...

Page 152 - If in AB the diameter of a circle produced a point C be taken, and therefrom a tangent as CD be drawn, and in the diameter a point E be taken fuch that AC : CB : : AE : EB ; then I fay ED being drawn will be perpendicular to the diameter AB.

Page 22 - Let there be two circles having the fame centre, and let the circumference of one of the circles be divided into any number of equal parts, and from the points of...

Page 160 - T = \pc r \2~Y? 13. If the spring's elastic force, when wound through the given angle or arc a= 90° from the quiescent position, be = p ; the weight of the balance, and the parts which vibrate with it = w, the distance of the centre of gyration from the...

Page 147 - D any line be drawn to meet the circle in E and F, and CE, CF be joined ; then I say EC : CF : : ED : DF. PRoP. V. If the base BC of a triangle be bisected in D, and through the vertex A a parallel thereto be drawn, and from D a perpendicular to BC be drawn to meet the parallel in E, and through D any line be drawn to meet AB, AC in F and G, and EF, EG be joined ; then I say EF : EG : : FD : DG. PRoP. VI. If in the line AB be taken two points C and again in h', E', and TG to T...

Page 148 - CD, v/hich meets the diameter AB in F and the circumference in G ; then I fay that DC : CE : : DF : FG. PROP. X. If in AB the diameter of a circle two points C and D be taken fuch that AC : CB :: AD : DB, and through the centre E a perpendicular to AB be drawn, and from C a line be drawn to meet the fame in F, and if through D any line DG be drawn to meet the circle in G and H, and from the point G be drawn GK the fame fide of DG as F is of the diameter AB to make the angle DGK equal to the angle...

Page 216 - ... figure, for since AE . AF = AB* = AC . AD, the points C, D, E, F are in a circle, therefore the external angle ACE = EFD and = DHG ; .-. EFD being = and from A and B two lines be drawn to touch the circle in C and D, and let the square of AB be equal to the sum of the squares of AC and BD, and from A any line be drawn to meet the circle in E and F, and BE, BF be drawn meeting the circle again in G and H ; the points A, G, H, will be in a right line.